Question: Find the smallest solution to the equation \[\frac{2x}{x-2} + \frac{2x^2-24}{x} = 11.\]
Multiplying both sides by $x(x-2),$ we get \[2x^2 + (2x^2-24)(x-2) = 11x(x-2),\]which simplifies to \[2x^3 - 13x^2 - 2x + 48 = 0.\]Looking for rational roots to the equation, we see that $x=6$ is a solution. Performing the polynomial division, we get \[2x^3 - 13x^2 - 2x + 48 = (x-6)(2x^2-x-8) = 0,\]so either $x = 6$ or $2x^2 - x - 8 =0.$ The latter quadratic has solutions \[x = \frac{1 \pm \sqrt{65}}{4},\]so the smallest root of the original equation is $x = \boxed{\frac{1-\sqrt{65}}{4}}.$